Question

Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function

Short answer

The Maclaurin series for the given function is \(\sum\limits_{n = 0}^\infty {\frac{{\left( {1 + 2{{( - 1)}^n}} \right)}}{{n!}}} {x^n}\)

Step-by-step solution

Step 1: Applied the Maclaurin series

By replacing the Maclaurin series function of given function

\(\begin{aligned}{l}f(x) = {e^x} + 2{e^{ - x}}\\{e^x} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( x \right)}^n}}}{{n!}}} \end{aligned}\)

Replaced by in maclaurin series

Step 2: Maclaurin series function of the given function

Maclaurin series function of the given function

\(\begin{aligned}{l}{e^x} + 2{e^{ - x}} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} + 2\sum\limits_{n = 0}^\infty {\frac{{{{( - x)}^n}}}{{n!}}} \\{e^x} + 2{e^{ - x}} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} + 2\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \cdot \frac{{{x^n}}}{{n!}}\\{e^x} + 2{e^{ - x}} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \left( {1 + 2{{( - 1)}^n}} \right)\\ = \sum\limits_{n = 0}^\infty {\frac{{\left( {1 + 2{{( - 1)}^n}} \right)}}{{n!}}} {x^n}\end{aligned}\)